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Laplace transformation techniques in operationalcalculusLewis WootenAtlanta University
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LAPLACE TRANSFORMATION TECHNIQUES IN
OPERATIONAL CALCULUS
A THESIS
SUBMITTED TO THE FACULTY OF ATLANTA UNIVERSITY
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR
THE DEGREE OF MASTER OF SCIENCE
BY
LEWIS WOOTEN
DEPARTMENT OF MATHEMATICAL SCIENCES
ATLANTA, GEORGIA
DECEMBER 1979
ABSTRACT
MATHEMATICS
WOOTEN, LEWIS, Atlanta University, 1979
Laplace Transformation Techniques in Operational Calculus
Advisor: Dr. Benjamin Martin
Master of Science degree conferred December 20, 1979
Thesis Dated: September 1979
This paper is concerned with the application of the theory of Opera
tional Calculus based on the Laplace transformation to problems frequently
encountered in Applied Mathematics. Beginning with a minimum of the theory
of the Laplace transform, and applying this basic theory of ordinary dif
ferential equations with both constant and variable coefficients. The pro
cess for finding the solution to partial differential equations with x, and
t as independent variables is illustrated. Additional topics are extended
to include the evaluation of definite integrals, nonlinear differential
equations and Volterra's integral equations with different kernels are also
discussed.
CONTENTS
INTRODUCTION iv-
CHAPTER PaSe
I. DEFINITION AND FUNDAMENTAL THEOREMS 1
Definition of the Laplace Transformation 1
Fundamental Theorems 3Inverse Transform • 9
II. APPLICATION TO ORDINARY LINEAR DIFFERENTIAL EQUATIONS
WITH CONSTANT COEFFICIENTS 12
Single Differential Equation 12Simultaneous Differential Equations 19
III. APPLICATION TO ORDINARY DIFFERENTIAL EQUATIONS WITH
VARIABLES COEFFICIENTS • 24
IV. APPLICATION TO PARTIAL DIFFERENTIAL EQUATIONS 28
V. EVALUATION OF DEFINITE INTEGRALS 40
VI.• APPLICATION TO ORDINARY NONLINEAR DIFFERENTIAL EQUATIONS 47
Solution by Nonlinear Integral Equations 48Solution by Power Series
VII. APPLICATION TO INTEGRAL EQUATIONS OF THE CONVOLUTION TYPE 61
Integral Equations of the Convolution Type 62Abel's Integral Equations •
APPENDICES • 71
BIBLIOGRAPHY ' 77
ii
List of Tables
Table PaSe
Tables of Properties 71
Tables of Transforms ' •*
iii
INTRODUCTION
The Laplace transformation was first proposed by Pierre Simon
Laplace, a French mathematician and one of the most prolific mathemati
cian of the 18th century. Very little is known of his early years for
when he become distinguished, he had the pettiness to hold himself aloft
from both his relatives and from those who had assisted him. Laplace was
more of an applied mathematician, having done the majority of his work in
mathematical physics, mechanics, and astronomy. During the years 1784-
1787, he produced some numerous works of exceptional power. Prominent
amount of these is one, which was read in 1784, and reprinted in the third
volume of the Mechanique Celeste, in which he introduced the Laplace inte
gral.
The term "operational method" implies a procedure of solving dif
ferential equations whereby the boundary or initial conditions are auto
matically satisfied in the course of the solution. Much.of the interest
in operational method was stimulated by Oliver Heaviside (1850-1925) who
developed its earlier concepts and applied them successfully to problems
dealing with almost every phase of physics and applied mathematics. In
spite of his notable contributions, Heaviside's development of the opera
tional calculus was largely empirical and lacking in mathematical rigor.
The operational method was placed on a sound mathematical founda
tion through the. efforts of many men. Bormwich and Wagner (1916) were
among the first to justify Heaviside's work on the basic of contour inte
gration. Carson followed by formulating the operational calculus on the
iv
basic of the infinite integral of the Laplace type. The methods of Carson
and Bromwich were linked together by Levy amd March as two phases of the
more general approach. Van der Pol, Doetsch, and others contributed to
summarizing the earlier works into a procedure of solution presently known
as the operational method of Laplace transformation.
Problems involving ordinary differential equations can be solved
operationally by an elementary knowledge of the Laplace transformation,
whereas other problems leading to partial differential equations required
some knowledge of the complex variable theory for thorough understanding.
The operational method of Laplace transformation offers a very
powerful technique for the fields of applied mathematics. In contrast to
the classical method, which requires the general solution to be fitted to
the initial or boundary conditions, these conditions are automatically in
corporated in the operational solution for any arbitrary or prescribed
excitation. Solutions for impulsive types of excitation and excitation of
arbitrary nature can be concisely written operationally. In some cases, it
is possible to determine the behavior of the system merely by examining the
operational equation without actually carrying out the solution.
The activity stimulated by Heaviside's method has been aptly sum
marized by the eminent British mathematician, E. T. Whittaker* in the
following statement:
Looking back on the controversy after thirty years, we should nowplace the operational calculus with Poincare's discovery of auto-morphic functions and Ricci's discovery of tensor calculus as thethree most important mathematical advances of the last quarter ofthe nineteenth century. Applications, extensions, and justificationof it constitute a considerable part of the mathematical activity of
today.
*Whittaker, E. J.: Oliver Heaviside, Bulletin of the Calcutta Mathe
matical Society, Volume 20, p. 199.
CHAPTER I
DEFINITION AND FUNDAMENTAL THEOREMS
1*1 Definition of the Laplace Transform
Definition 1.1_L Let f (t) be a function of t specified for t *■ 0. Then the
Laplace transform of f(t) denoted by f(t), is defined as
e"stf(t) dt* (l.l.D
where the parameter s is complex.T
The Laplace transform of f(t) is said to exist if the integral
(l.l.l) converges for some value of s; otherwise it does not exist. The
sufficient conditions under which the Laplace transfaorm exist are that
the function f(t) in (1.1.1) is sectionally continuous in every finite in
terval 0< t<N, and the function must be of exponential order for t > N.
The new function F(s) is called the Laplace transform, or the
image, of the original function f(t). Whenever it is convenient to do so,
we shall denote the original function f(t) in lower-case letters and its
transform by the same letter in upper-case. But other notations that
distinguish between functions and their transforms are sometimes prefer
able; for example,
<£(s) = L f(t) or y(s) = L f(t)
*&is the symbol normally used to denote the Laplace transformo£ a function, but for printing purposes we shall use the letter L.
fWe assume at the present time the parameter is real.
1
A function is called sectionally continuous or piecewise con
tinuous in an interval a < t< b if it is such that the interval can be
divided into a finite number of subintervals in which the function is con
tinuous and has finite right and left hand limits.
The unit step function, is defined as
10 when < t < k • (1.1.2)
1 when t > k
is an example of a function that is sectionally continuous on the interval
0 < t < N for every positive number N.
A function f(t) is of exponential order as t tends to infinity,
provided there is some constant M > 0 such that the product
je~atf(t)| < M or Jf(t)| < Me"" (1.1.3)
we say that f(t) is a function of exponential order as t-»«>or, briefly,
is of exponential order.
The function U(t) above, as well as the function t , are of
order of ** ast*arfor any positive a ; in fact, for the first function
and, when n = 0, for the second, we may write a = 0. The function e t of
exponential order (C > 2); but the function et2 is not of exponential
order.
Theorem 1.1: Sufficient Conditions for the Existence of the Laplace
Transform: If f(t) is sectionally continuous in evey finite interval
0 < t < N and exponential order a for t > N, then its Laplace transform
F(s) exist for all s > a .
Proof: We have for any positive number N,
-stf(t) dt =rvstf(t)dt +/\rstf(t) dt U.I.*)
Since f(t) is sectionally continuous its every finite subinterval
0 < t< N, the first integral on the right exist. Also, the second in
tegral in the right exist, since f(t) is of exponential order a for
t > N. To see this we have only to observe that in such case
«<t)| dt
. a
< / e~BtVlet dt = Mo T^T (1.1.6)
Thus the Laplace transform exists for s > a .
1.2 Fundamental Theorems of the Laplace Transform
We will now consider some of the very powerful and useful general
theorems concerning operations of the Laplace transform. These theorems
are of great utility in the solution of differential equations, evaluation
of integrals, and other procedures of applied mathematics.
In the following list of theorems, we assume, unless otherwise
stated, that all functions satisfy the conditions of Theorem 1.1 so that
their Laplace thansform exist.
Theorem 2.1: The Laplace transform of a constant is the same constant
divided by s, that is
Lk =¥ d.2.1)
Proof: To prove this, we have from the definition of the Laplace transform
r60 -<?t -st~\P ,Lk =/e kdt -1^ e_J .i (1.2.2)
The integral vanishes at the upper limit since by hypothesis
Res > 0.
Theorem 2.2; L k0(t) = k L H(t), where k is a constant.
Proof: We can prove this in the following manner
(1.2.3)
L kH(t) "StkH(t) dt = k/ e"StH(t) dtJ o
= kL H(t) (1.2.4)
Theorem 2.3: Linearity Property
If c and c. are only constants while f^(t) and f2(t) are func
tions with the Laplace transforms F^s) and F2(s) respectively, then
c2f2(t)]2f2 f2(t)
(1.2.5)
The result is easily extended to more than two functions. The
proof follows directly from Theorem 2.2, and is easy enough, and will not
be given there. Because of the property of L expressed in this theorem,
we say L is a linear operator or that it has the linear property.
Theorem 2.4: Laplace Transform of Derivatives
If f(t) is continuous and has a derivative feed which Laplace
transformable
If L f(t) - F(s),
L f'(t) = sF(s) - f(0) (1.2.6)
Proof: To prove it, we have
Lf'(t) =rooe"StfI(t) dt (1.2.7)
Integrating by parts we
to
= f(t)e
t
lim f(t)e
o
-
-stf(t) dt
I sL f(t)
sF(s) - f(0) (1.2.8)
This theorem is very useful in solving differential equations
with constant coefficients. This method is useful in finding the Laplace
transform of derivative without integrating, when the transform of the
function is known.
Theorem 2.5: If f(t) is continuous and has derivatives of order 1, 2, 3,
..., n which are Laplace transformable, then
Lf(n)(t) =snF(s) - Y fW (0)."-W (1.2.9)
(k) kwhere F(s) is the Laplace transform of f(t) and f (0) = d_f
dtk
evaluated at t = 0.
This theorem is an extension of Theorem 2.4. A proof will not
be given here.
By repeated application of Theorem 2.4 the required results can
be obtained. Therefore it follows that
L f'(t) = sF(s) - f(0) (1.2.10a)
L f"(t) = s2F(s) - sf(0) - f'(0) (1.2.10b)
L f(3)(t) = s3F(s) - s2f(0) - sf'(O) - f"(0) (1.2.10c)
and etc. These expressions are useful in transforming differential
equations.
Theorem 2.6: If L f(t) = F(s), then
f f(u) du = F(s) (1.2.11)o
Proof: Since
L
o
/flu) =/>°VSt fhin) du (1.2.11b)J n J n J O
Now integrating by parts yields
L f f(u) du =le"Stf f(u) du + I F(s) (1.2.11c)J o J O J O S
s .
Now at the upper limit the first term vanishes because of the exponential
function. At the lower limit the first terms vanishes because of the
definite integral. Hence only the second term is left, and it is the
result stated in Theorem 2.6.
Theorem 2.7: Multiplication by t .
If L f(t) = F(s)
,n (n)
L tnf(t) = (-l)n*-T(s) = (-l)nF (s) (1.2.12)ds11
where n = 1, 2, 3
Proof: We have
F(s) = fVStf(t) dt d.2.13)
Then by the Leibnitz's rule for differential under the integral sign,
*df m
ds
F'(s) = d
ds
- rJo
Jo~stf(t) dt =/*°D
Jo
-te"Stf(t) dt
/:•-st
i tf(t) dt
-st
d e fO
ds
:) dt
- L tf(t) (1.2.14)
Thus L tf(t) = -df_ = -f*(s)ds
which proves the theorem for n = 1.
*We assume here that the Leibnitz's rule can be applied.
To establish the theorem in general, we use matheraatical induction.
Assume the theorem true for n = k, that is assume
" tkf(t) dt = (-l)Vk)(s) (1.2.15)
Then
' e"St tkf(t) dt = (-l)kFk+1)(s) (1.2.16d_ f
dsjoo
or by Leibnitz's rule,
e t f(t) dt = (-1) F (s) (1.2.17;
that is
r°°e-st t k+lf(t) dt = (.D^p^^Cs) (1.2.18)J o
It follows that is (1.2.15) is true, i.e., if the theorem holds for n = k,
then (1.2.18) is true, i.e., the theorem holds for n = k +1. But by
(1.2.14) the theorem is true for n = 1. Hence it is true for n = 1 + 1 = 2
and n = 1 + 1 = 3, etc., and this for all positive integer values of n.
Theorem 2.8: Differentiation with respect to a second independent variable.
Consider the following function of two independent variables:
f = f (x,t)
If L f(x,t) = F(x,s) (1.2.19)
then ^ 8f(s:0 = iFOy) (1>2>20)
Proof: the proof of this theorem follows directly from the definition of
the Laplace transform:
L 9f(x,t) = f e"Staf(x,t) dt (1.2.21)*x JO Jx
6
Since the variable x is not a variable of integration, then the order of
differentiation and integration may be interchanged to give
Lf(x,t) = 2L f e"St tl*^ dt (1.2.22)
*x
which completes the proof.
Theorem 2.9: Convolution Theorem*.
L fx(t) = F^s) (1.2.23)
L fx(t) = F2(s) (1.2.24)
Then the theorem states that
hf'f. (u)f (t-u) du = ht f2(u)f1(t-u)du = F1F2 (1.2.25)
The proof of Theorem 2.9 will not be given here, and we will list the
following theorem without their proofs.
Theorem 2.10: First Shifting Theorem
If L f(t) - f(s) ! - (1.2.26)
Then L eatf(t) = F(s - a) U-2-2?)
Theorem 2.11: Second Shifting Theorem
If L f(t) = F(s) and
f-F(t -a) t >
lo t < ag(t)
L g(t) = e-aSF(s)
*This theorem is known in the literature as the Faltung or convolution theorem. In the older literature of operational calculus it is called
the superposition theorem.
1.3 Inverse Transform
Definition 3.1: Let the symbol l"1 F(s) denote a function whose Laplace
transform is F(s). Thus
L f(t) = F(s) (1.3.1)
then
f(t) = L'1 F(s) (1.3.2)
This correspondence between function F(s) and f(t) is called
the inverse Laplace transformation, f(t) being the inverse transform of
F(s). The operator L"1 is a linear operator or that it has the linearity
property.
Since the Laplace transform of a null function >7(t) is zero,
that is
f net) dt-o (1-3-3)o
it is clear that L f(t) = F(s) then also [f(t) + tft)] = F(s).
From this it follows that we can have two different functions with the
same Laplace transform.
Example: The two functions
and
e~3t otherwise U.3.5)
have the same Laplace transform, that is l/(s + 3).
If we allow null functions, we see that the inverse Laplace
transform is not unique. It is unique, if we disallow null functions
(which do not generally arise in cases of physical problems).
10
Theorem 3.1; Lerch's Theorem.
If we restrict ourselves to functions f(t) which are sectionally
continuous in every finite interval 0 <_ t<_ N and of exponential order for
t > N, then the inverse Laplace transform of F(s), that is L F(s) = f(t)
is unique.
We shall always assume such uniqueness unless otherwise stated.
The most obvious way of finding the inverse transform of a given
function of s consist of reading the result from a table of transforms.
But we shall take up methods of obtaining inverse transforms of certain
combinations and modifications of function of s, as well as methods of
resolving such functions into those listed in the tables. With the aid of
such procedures, we shall be able to make much use of the Laplace transfor
mation. In addition, there are explicit formulas for L F(s). The most
useful of these formulas involves an integral in the complex plane. To use
this integral, we must let s be a complex Variable and we must be prepared
to employ some theorems in the theory of functions of a complex variable.
Definition 3.2: If F(s) = L f(t), then L"1 F(s) is given by
1 a+iaj
f(t) - 2±iT f eStF(s) ds, t > 0 (1.3.6)
^ a-ia>
and f(t) = 0 for t < 0.
This result is called the complex inversion integral or formula.
It is also known as Bromwich's integral formula. The result provides a
direct means for obtaining the inverse Laplace transformation of a given
function F(s).
The integration of (1.3.6) is performed along a line s = a in
the complex plane where s = x + iy. The real number a is chosen so that
11
s = a lies to the right of all singularities (poles, branch points or
essential singularities) but is otherwise arbitrary.
CHAPTER II
APPLICATION TO ORDINARY LINEAR DIFFERENTIAL
EQUATIONS WITH CONSTANT COEFFICIENTS
The application of the Laplace transformation to solutions of
linear ordinary differential equations with constant coefficients, or
systems of such equations is well known. Such problems can, of course, be
solved by methods studied in a first course in differential equations.
We shall later on, solve more difficult problems, especially those in par
tial differential equations.
2.1 Single Differential Equations
In this section we will consider from a practical point of view,
a physical system which is characterized by a single time function y(t),
which satisfied a single differential equation. When more than one time
function exists in a physical system and they satisfy more than one dif
ferential equation in which all or some of those functions occur (simul
taneous differential equations).
We can exhibit the method of solution used here in a clearly
arranged scheme, to which we always return when we employ the Laplace
transformation of functional equations.
12
13
SCHEME
Original!space:!
Differential equation
initial conditions
Solution
Laplace transformation
Inverse
Laplace transformation
Image
space:! Algebraic equation Solution
Explanation: Instead of solving the given differential equation
with given initial conditions directly, we make a detour across into the
iiaage space. We go from the original equation to the image equation (an
albegraic equation) by the Laplace transformation, we solve this and then
translate the solution back to the original space with the help of the in
version formula or the tables of transforms.
Hence the image function Y(s) of the desired time function y(t)
can be found by applying the Laplace transformation and we require only to
determine its corresponding original function. For this purpose we could
use the complex inversion formula. However, we wish to avoid this as
much as possible and so we proceed in the same way as when we meet an
integral in the process of calculation. We do not evaluate the integral
from its definition as the limit of a sum but we consult a table of inte
grals; when we do not find the required result immediately, we decompose
and reshape the integrand so that we reach a known integral. Analogously,
in our case we consult the attached Tables of Transforms and see whether
the original function of the image function under consideration is written
14
down, or whether perhaps by using our 'grammatical rules,' see if the
function can be built up from the functions in the tables.
There are advantages for using the Operational method (Laplace
transform) as opposed to the classical methods. In the classical method
we establish first a 'general' solution which the constant must be made to
fit the initial values. This necessitates the solution of an additional
system of linear equations with an unknown (for n >3 this is fairly tedious),
By contrast, the Laplace transform considers these initial values from the
beginning and introduces them automatically into the solution. Because of
this their influence is clear from the start. Therefore, the method is
particularly suitable for the initial valued problems. The frequent case
of vanishing initial values, which is not any simpler in the classical
method still requires the solution of the previously mentioned system of
linear equations, is solved very simply by the Laplace transformation.
While the classical method we first solve the homogeneous equation
and then the inhomogeneous equation by variation of parameters, in using
the Laplace transformation we can immediately solve the inhomogeneous equa
tion, which is practice the more important.
We will illustrate the procedure by considering the general equa
tion with constant coefficients;
x(n) (t) = alX(n-l Vl n
We introduce the following transforms:
y(s) = L x(t) = f e~Stx(t) dt (2-2)•'O
Applying the Laplace transform to the differential equation, we have
15
(sny(s) -
- sn"3x'(0) - ... sx(n"3)(0) - x(n"2)(0)
- x(0) + any(s) + f(s) . (2.3)
If we use the notation
g(B) = (x(n-1}(0) + sx(n-2
a (x(n"2)(0) + sx(n"3)(0) + . . . sn-2x(0) +
... a .x(0) (2-4>n-1
New we can write equation (2.1.4) in the form
, n+ a.811"1* . • • a)y(s) + g(s) = f(s) (2.5)(si n
If we let
3n+a,sn-1+ * * ' an (2-6)j, (s) = s + a'n 1
Then the image equation can be written as
y(s) = sSfD+Jlsl - L *<*> (2*7)L (s) L (s)n n
To obtain the solution of the differential Eq. (2.1.1) we must obtain in
some manner the inverse transform of y(s), and we would then have the
desired solution. The procedure in such a case is to decompose the expres
sion g(s)/Ln(s) and f(s)/Ln(s) into partial fraction, examine the table of
transforms, and obtain the appropriate inverse transforms.
We will consider some examples to illustrate the procedure.
16
Example 1.1 Find the general solution of the differential equation
y"(t) + k2y(t) = 0
y(0) = A, and y'(0) = B (2.1.2)
If we let
Y(s) = L y(t) -JQ eSty(t) dt (2.1.3)
then applying the Laplace transform to both members of Eq. (2.1.1), and
making use of the differentiation theorem, we obtain the following equa
tion:
s2Y(s) - sy(0) - y'(0) + k2Y(s) = O (2.1.4)
If the function ;<(t) satisfies the initial conditions, then
s2Y(s) - sA - B + k2Y(s) = 0 (2.1.5)
which is a simple algebraic equation. Its solution is clearly
Y(s) = A |— 1 + I I
Now inverting Eq. (2.1.6), we find
y(t) = Acos kt + B. sin kt
k
= Acos kt + B'sin kt (2.1.7)
where B1 = B/k, and A and B' are arbitrary constants.
To verify our formal result given by Eq. (2.1.7), we need only find
y"(t) from that equation and substitute in Eq. (2.1.1) to see that the
differential euqation is satisfied regardless of the value of A and B'.
17
Example 2.2 Find the solution of the differential equation
y"(t) + 2y'(t) + (5y(t) = e^sin t (2.1.8)
subject to the following initial conditions
y(0) = 0, and y'(0) = 1 (2.1.9)
Applying the Laplace transform to both members of the differential
equation, and letting Y(s) denote the transform of y(t), we obtain
s2Y(s) - sy(0) - y'(0) + 2 sY(s) - y(0) = 5Y(s) -
(2.1.10)
(s + I)2 + 1
Considering the initial conditions, we have
s2Y(s) - 1 + 2sY(s) + 5Y(s) = l/((s + I)2 + 1)
Hence
(s2 -r 2s + 5)Y(s) = l/(s2 + 2s + 2) (2.1.11)
Therefore, the solution of the image equation is
Y(s) = s + 2s + 3 (2.1.12)
(s2 + 2s + 2) (s + 2s + 5)
The rational function (2.1.12) can be expanded into fractions in
which the denominators contain only linear factors.
s2 + 2s + 3 = As + B = Cs + D (2.1.13)
(I2 + 2s + s)(s2 + 2s = 5) s2 + 2s + 2 s2 + 2s + 5
Solving for A, B. C and D from the partial fraction expansion (2.1.13),
we find that
A = 0, B = 1/3, C - 0, and D = 2/3
18
Now we have
Y(s)
3(s2 + 2s + 2) 3(s2 + 2s +5)
or
Y(s) - 1—~ + ^ (2.1.14)3((s + IV + 1) 3((s = IV + 4)
Hence
1 + 2 I
+ I)2 + 1) 3«s + I)2 + 4)Jy(t) = IT1 |"_ 1 + 2 1 (2.1.15)
L3((s
Therefore, we have
^ -re si^-t +• -re sin 2t3 j
or
y(t) - 1 "'(sin t + sin 2t) (2.1.16)
Example 1.3 Find the general solution of the following differential
(2.1.17)
0) C1, and y(0) C2 (2.1.18).
If we let
Y(s) = L y(t) = J e"Sty(t) dt (2.1.19)
then taking the Laplace transform of Eq. (2.1.17), we have
s2Y(s) - sy(0) -y'(0) + a2Y(s) = F(s) (2.1.20)
equation
subject
•
•
y"(t) H
to the
y(0) =
- a2y(t) =
conditions
C^ and y'
t(t)
(0)
19
where
F(s) = L f(t)
Substituting the initial condition (2.1.18) in Eq. (2.1.20) and solving,
we have
sCl C2 F(s)Y(s) = i_ + Z + nSJ (2.1.21)
2,2 2^2 2, '2s+a s+a s+a
Inverting Eq. (2.1.21), we have
y(t) + C cosh at + _2 sinh at + 1 / sinh a(t - u)f(u) du1 o J n
or
y(t) = C cosh at + C ' sinh at + 1 C sinh a(t - u)f(u) du■*■ a J oa J o
(2.1.22)
where C^ = C2/a.
2.,1 Simultaneous Differential Equations
As has already been shown, the calculations involved in solution
of a single differential equation with order greater than three are much
simpler by the Laplace transform method than by the classical method. How
ever, the method shows its full power in the solution of systems of several
differential equations where it leads to greater insight and fewer calcula
tions than the classical method, which in reality is not practicable at all.
For the sake of simplicity, consider the system of three differential
equations of the first order. In these, we write down all terms that theo
retically can appear, although usually a number of these terms are absent,
so that their coefficients have then to be set equal to 0. The system to
be considered is
20
(a21y'. +b12yi) + (a22yI2 + b22y2) ^a^y' + b23y3} = f2(t)
(a31yl + b31yl> + (a32fc2 + b32y2) + (a33y3 + b33y3) = f3(t)
Taking the Laplace transform of the system yields
ai3 (SY3 -y3(0)) + b13Y3 = F1(S)
a21(sY1-y1(0))
331
"y3
-y
-y
(0))
2<°>)
3(0))
(2.2.2)
With the abbreviation
a s + b = o (s) (2.2.3)aiks + Dik Pikk ;
We can write these equations in the form
PllY1 + Pl2 Y2 + P13Y3 = Fl + aHyl(0) + a12y2(0) + al3y3(0)
P21Y1 + P22Y2 + P23Y3 = F2 + a21yl(0) + a22y2(0)"+ a^
p31Yl + P32Y2 + P33Y3 + F3 + a31yl(0) + a32y2(0) + a33y3(0)
If the differential equations were not of first but of second order, then
the p (s) would be polynomials not of the first but of the second degree
and the value y'(0) ... would appear on the right-hand side in addition to
21
to the value of y(0) .... In practice, however, the equation would be
similar; they form a system of linear algebraic equations for the unknown
Y., Y_, Y»; theoretically such a system is solved most elegantly by
Crammer's rule with determinants, but in practice it is sometimes better
to use the successive elimination of the unknowns or (with a great number
of equations) one of the several different methods developed for solving
such a system.
On the right-hand, side of the equations there are the image func
tions F.(s) of the input function f.(t), and numerical constants which de
pend on the initial values; these we combine as follows:
ailyl(0) + aily2(0) + ai3y3(0) = ri (2.2.5)
Let D(s) be the determinant of the system which is built from the P-k(s)
and which is in general a polynomial of the third degree in a, and we
obtain by Cramer's rule
'I"*
rl P12 P13"1
2 + r2 P22 P23
LF3 + r3 p32 p33j
^1
Y3i
21
P12F1
P21 P22 F2
'32 X3
P13"
23 (2.2.6)
22
This solution satisfies the system of differential equations and
the initial condition, provided that the determinant of the coefficients
of the highest powers of the system does not vanish, that is
det [a±k] * o (2.2.7)
and this is called the normal case. In this section we are making the
restrictions to only the normal case.
The procedure is illustrated by the following example:
Example 2.1 Solve the following system of differential equations.
x'(t) = 2x(t) - 3y(t) (2.2.8)
y'(t) = y(t) - 2x(t)
aubject to the initial conditions
x = b)> t - o <2-2-9>
y - 3J
Taking the Laplace transform, where
y*00 -ste x(t) dt
o
. ■ , (2.2.10)and
x " r° -stY(s) - L y(t) -
we have
sX(s) - x(0) = 2X(s) - 3Y(s) (2.2.12)
sY(s) - y(0) = Y(s) - 2X(s)
substituting the initial conditions and simplifying, we have
(s-2)X(s) ♦ 3Y(s) - 8 (2
2X(s) + (s - l)lf(s) = 3
23
Solving Eqs. (2.2.13) simultaneously by Cramer's rule, we find that
8
_■ 3
f-
2
s
2
s
3
-
3
-
1
-
.1
8s- 17 8s - 17
s -3s -4 (s + l)(s - 4)
5_ 3
s + 1 s - 4
s - 2 8
2 33s - 22 3s - 22
s - 2 3
2 s - 1
3Z- 3s - 4 (s
s '+ 1 s-4
(2.2.15)
Then inverting the above equations, we have
X(t) = L"1 X(s) =
y(t) - L"1 Y(s) = 22e"4t
(2.2.16)
CHAPTER III
APPLICATION TO ORDINARY DIFFERENTIAL EQUATIONS
WITH VARIABLES COEFFICIENTS
It is sometimes possible, by the introduction of the Laplace trans
formation, to transform certain linear differential equation with vari
ables coefficients to other equations that may be integrated readily. A
linear differential equation in y(t) whose coefficients are polynomials
in t transforms into a linear differential equation in y(s) whose coeffi
cients are polynomials in s. In case the transformed equation is simpler
than the original, the transformation may enable us to find the solution
of the original equation.
If the coefficients are polynomials of the first degree, the trans
formed equation is a linear equation of the first order, whose solution
can be written in terms of an integral. To find the solution of the
original equation, however, the inverse transform of the solution must be
obtained.
The procedure will be illustrated by the following examples:
Example 1. Find the solution of the differential equation
y"(t) + ty'(t) - y(t) = 0 (3-D
subject to the following initial conditions:
v = Ct = 0 (3.2)
y'
24
25
We have seen that
L tny(t) = (-1)° _dnL y(t) = (-1)Vn) (a) (3-3)ds
and therefore we can write the transform of the product of t and any
derivative of y(t) in terms of y(s).
To transform Eq. (3.1), we let
Y(s) - r" e"8ty(y) dt .(3.4)J
- r
J
then the transformed equation is
s2Y(s) - sy(0) - y'(0) - _d gy(s) _ y(Q) _ v(s) = Ods
(3.5)
Considering the initial conditions, and simplifying we have
Y'(«) + Y(s) (I - s) =-- (3.6)
An integrating factor is
2
exp f f(2 - s) ds] = exp (21n s - hs )I" ff'Z
2 -^ (3.7)= s e
so the equation can be written as
d 2 -h2 - - ks2e~hS ^ (3-8)ds
or
ds e y(s) = -se ds
integrating, we have
"* ds (3-9)
where C is the constant of integration
Y(s) = J^ + C %g2 (3.11)
26
To determine C, note that by series expansion
Y(s) - -„+ —-(1 + -2+ 1 4+...) (3.12)
s s 2 IIsthen
Y(s)=i+£+ C(I+(I2 + ...) (3.13)
s2 2 8S
Then since l"1 sK = 0, where k = 0, 1, 2, ..., we obtain after inverting
y(t) = (1 + C)t (3.14)
But from Eq. (3.2) we have y'(0) = 1, so calculate C = 0, then the required
solution is
y(t) = t (3-15)
Example 2; Solve Bessel's equation of order zero.
ty"(t) + y'(t) + ty(t) = 0 (3.16)
under the conditions that y(0) = 1 and y(t) and its derivatives have trans
forms .
The point t = 0 is a singular point of this differential euqation
such that one of the solution is a function that behaves like the natural
log of t near the singular point, and the Laplace transform of the deri
vatives of the function does not exist.
Applying the Laplace transform to the equation we have
_JL s2Y(s) - sy(0) - y'(0) + sY(s) - y(0) - ^(s) = 0 (3.17)ds
and substituting the initial conditions we have
(s2 + l)Y'(s) + sY(s) = 0 (3.18)
This is a first order differential equation and can be solved by separa
tion of variables. Separating the variables, we have
rs + 1
27
and integrating gives
In Y = -Jain (s + 1) + (3.20)
where c is the constant of integration. Solving we find that
Y(s) = c/(s2 + l)h (3.21)
consulting the tables on transforms, we find the solution to the original
equation
y(t) = CJ (t) (3.22)
If the function is to satisfy the condition y(0) = 1, then we
have C = 1, and the solution can be written as
y(t) = J (t) (3.23)
where J (t) is Bessel's function of the first kind of order zero:o
Jo(t)CD
n=o
(-1)n
(n!)' (l)2n
(3.24)
•CHAPTER IV
APPLICATION TO PARTIAL DIFFERENTIAL EQUATIONS
In Chapter 2 we found that problems leading to ordinary differen
tial equations could be solved easily by the use of the Laplace trans
formation. In the study of partial differential equations it will appear
that it is still easy to find the solution by the Laplace transform, but,
since this usually involves more complicated functions of s than those
considered in Chapter 2, it is rather more difficult to derive the solution
by it.
In partial differential equations the unknowns are functions of
several variables, we consider here the case of two variables which we
denote by x and t; the unknown function is denoted by u(x,t). In a partial
differential equation a certian domain in the xt-plane is given at the .out
set, and it is within this domain that the unknown is to be determined.
For the equations considered here, we assume that t varies in the one-sided
infinite interval 0 < t <«,/ and x varies in a finite or infinite interval,
that the basic region of the xt-plane is a semi-infinite script, a quadrant
or a half-plane according as x varies in a finite, one-sided or two sided-
infinite interval.
If we wish to solve partial differential equations by using the
Laplace transformation, we must apply the Laplace transformation to the
function u(x,t) and to the derivatives which occur. Because the transfor
mation represents an integration with respect to a single variable, we
28
29
must undertake it relative to one of the variables in u, while the ether
is not involved. The variable to which the transformation is applied is t,
which we have assumed from the first to vary be between 0 andao, because
this is the interval over which the Laplace transform extends, the vari
able x is to be thought of as a constant. For each fixed value of x we
obtain a different transform which therefore depends not only (as previ
ously on s, but also on x; it is therefore a function U(x,s):
L u(x,t) = (* e"Stu(x,t) dt = U(x,s) (4.1)
If we wish to transform derivatives with respect to t, we can employ
Theorem (1.2.7), and here the variable x is kept constant. Then we have,
for example,
L u (x,t) = sU(x,s) - u(x,0), (A.2)
L u (x,t) = s2U(x,s) - xu(x,0) - ut(x,O) (4.3c. *-
To use our method for derivatives with respect to x it must be
assumed that the order of differentiation and integration in the Laplace
integral can be interchanged; for example
L V*-0 = dlL «(*.*> = s°<x'B> (4'4)
L uxt(x'° = dlL Ut(X>t) = ~& SU(X'S) " U(X'0) (4'5>Solution to one-dimensional boundary-value problems can be found
by the use of the Laplace transformation, by transforming the partial dif
ferential equation into an ordinary differential equation. The solution
is then obtained by solving the ordinary differential equation, and
*Where we have used the notation Ufc = du/dt
30
inverting by using the inversion formula or any other methods already
condidered.
By applying the Laplace transform twice we can find solutions to
two-dimensional problems, that is, we first transform the equation with
respect to one variable and then with respect to the remaining variable
and arrive at an ordinary differential equation. In such case the required
solution is obtained by double inversion. A similar technique can be
applied to three (or higher) dimensional problems. The process is some
times referred to as iterated Laplace transformation. Boundary-value prob
lems can sometime also be solved by both Fourier and Laplace transforms.
In the one-dimensional case, the method can be summed up in the
following scheme:
SCHEME
partial differential equation > solution
Original! + initial conditions »
space: | + boundary conditions
inverse Laplace
Laplace transformation transformation
1 ordinary differential equation
+ boundary conditions r> solution■
The most difficult part of the transformation method is normally
the determination of the original function of the function found by the
Laplace transform. If the present tables are not sufficient, then methods
of using the complex integral formula must be employed, or expanding the
image solution into a series of decreasing powers of s. But, for the
problems considered here, the present tables are sufficient.
31
The use of the Laplace transform in the solution of partial dif
ferential equations will now be illustrated by the following examples:
Example 1. Solve the partial differential equation
X(x,t) + xut(x,t) = 0 (4-6)
subject to the conditions
u(x,0) =0
u(0,t) = t (4'8)
Let
D(x.s) = r e-Stu(x,t) dt (4-9)
(4.10)
From Eq. (4.6) it follows that
JL(x.s) + x sU(x,s) - u(x,0) - 0
dx
Using the initial condition (4.7), Eq. (4.10) becomes
Uf(x,s) + xsU(x.s) = 0 (4'U)
subject to the condition
UCO.s) - Us1 <4-l2>
where U(0,s) is the transform of the boundary condition (4.8).
The partial differential equation has been reduced to the above
ordinary differential equation. We could solve Eq. (4.12) by the Laplace
transformation method, but, for the simple problems arising here, it is
just as easy to write the complementary function and particular integral
in the usual way.
Eq. (4.12) is a first order linear differential equation which can
be solved by separation of variables. Therefore, Eq. (4.12) can be written
in the form
dU=-sxdx •
U
32
Solving, we have
lnU = -Jssx2 + C (4.14)
Where C is the constant integration. Therefore, we can write
U - Ce~hs* (4.15)
To determine C, we consider the condition (4.12), and find that
I2 - C C4.16)s
Now we have the solution of the image equation
U(x,s) = I -*ssx2 (4-17)2 e
s
If we let
a = k>* C4.18)
and
f(s) = 1/s2
The Eq. (4.17) can be written in the following form
U(x,s) = f(s) e"aS (4.20)
Inverting Eq. (4.20), by consulting the table of transforms, we find that
u(x,t) = F(t - a)H(t - a) «
where F(t) = L*1 f(s) = t. There the solution can be written as
u(x,t) = (t - a)H(t - a)
or
uCS,t) = (t - *sx2)H(t - %x. )
where H(t) is the unit step function defined as in Chapter 1.
Example 2. Solve the boundary value problem
u (x,t) - 2u . (s,t) + u (x,t) = 0 (4.23)
subject to the following conditions
u(x,0) =0 (i
.33
ut(x,O) - 0
u(O,t) = 0
u(l,t) = 0 f(t)
Let
U(x,s) = L u(x,t)
Transofrming Eq. (4.23) we have
(4.25)
(4.26)
(4.27)
(4.28)
dxA sU(x,s) - u(x,0) + s2U(x,s) - su(x,O) - u (x,0)ax u
(4.29
Considering the initial condition, the above quation may be written as
U"(x,s) - 2sU'(x,s) + s2U(x,s) = 0 (4.30)
which is a second order ordinary linear differential equation that have
the following solution
U(x,s) - (Ax + B)eSX (4.31)
where A., and B are constant. To determine A and B we consider the boundary
conditions
U(0,s) = 0 (4.32)
U(l,s) = f(s) (4-33>
where conditions (4.32) and (4.33) are the transforms of (4.26) and (4.27)
respectively. Considering conditions (4.32), we have
B = 0
therefore
U(x,s) = Axe
Substituting conditions (4.33), we have
sx
F(s)
A
AeE
F(s)e-s (.35)
34
Therefore, the solution of the image equation is
U(x,s) = xF(s)es(x"1) «
If we let
a - (x - 1) (4.37)
then we can write Eq. (4.36) as
U(x,s) = xf(s)eaS " (4.38)
Inverting, we find
u(x,t) = xf(t + a)H(t - a) (4.39)
or
u(x,t) = xf(x + t - l)H(x + t - 1) (4.40)
where
xf(x + t - 1) t> X- 1
H(x + t - 1) - ' _0 t<X-1
Example 3. Let the temperature of the face of a semi-infinite solid x > 0
be prescribed by the function F(t) of time. If the initial temperature is
zero, the temperature function u(x,t) is the solution to the boundary value
problem.
ut(x,t) = ku (x,t) x >■ 0, t > 0 (4.41)t XX
u(x,0) --0 * ><> (4.42)
u(0,t) = f(t), limu(x.t) =0 t >0 (4.43)
Applying the Laplace transform to Eq. (4.41), we have
sU(x,s) - u(x,0) = kU"(x,s) (4-44)
where
U(x,s) = L u(x,t) = r° e-Stu(x,t) dt (4.45)r
J
35
Considering the initial conditions, and simplifying, we have
U"(x,s) - ~ U(x,s) = 0
Subject to the conditions
U(x,s) = F(s), and lim U(x,s) = 0
where
F(s) = L f(t)
The general solution od Eq. (4.46) is
U(x,s) = C^t . ~2"
We find 0= 0 from (4.47)
U(x,s) = C2e"X S
and from Eq. (4.47) we find c2 = F(s), so that
U(x,s) » F(s)e
From the values .of transforms we find that
-x/s/kexp
4kt
and with the aid of the convolution theorem, we find
D(x,t) - L"1 F(s)g(s) = r f(t -r)g(r) drJ o
where we have used the notiation
g(s) = arxVSTk
Now we have
t / 2 \u(x.t) = x f f(t - f ) expf - x )dr
"?W JQ r3/2
We now make the following substitution
(4.46)
(4.47)
(4.48)
(4.49)
(4.50)
(4.51)
(4.52)
(4.53)
(5.54)
(4.55)
(4.56)
36
and we calculate
2(4.57)
d - - x ,v (4.58)
and
.-3/2 = 8kXl (4.59)3
x
We can write Eq. (4.55) as
r x2 *■u(x,t) -2 r "
Since
2 ■? _.2
erfc =
Wlien the temperature of the surface is constant
the temperature within the solid is therefore
, „ "o *■> _ ,_^^X (4.62)u(x,t) =
therefore, Eq. (4.62) becomes
^ r P / X \ (4.64)u(x,t) = f erfc(—=-)
° \2Vkt /
Example 4t An infinitely long string having one end at x = 0 is initially
at rest on the x-axis. The end x = o undergoes a period!: transverse dis
placement given by A^in.t, t > 0. Find the displaced of any point on
the string at any time.
37
tAnsin
Fig. 4-1
If Y(x,t) is the transverse displacement of the string at any point x at
any time t, then the boundary-value problem is
(x,t) = a2y (x,t) x >0, t> 0 (4.65)
u(x,0) = yt(x,0) = 0 (4.66)
y(0,t) » A sin (o t, and |y(x,t)J < M (4.67)
where the last condition specifies that the displacement is bounded.
Taking the Laplace transform of Eq. (4.65), we find
s2Y(s,s) - sy(x,0) = a2Y"(x,s)
where
Y(x,s) = L y(x,t) =
/dt
>.68)
(4.69)
Considering the initial condition, and simplifying, Eq. (4.68) becomes
Y"(x,s) - s2
. a
subject to the conditions
Y(0,s) = Ao6>2 +,,2
s ***
Y(x,s) is bounded
The general solution of Eq. (4.70) is
Y(x,s) = C.eSX a + C e SX a
(4.70)
(4.71)
(4.72)
(4.73)
38
From the condition of boundedness, we must have C = 0. Then we have
(4.74)
(4.75)
From condition (4.71), we find that
J2 2
Therefore,
Y(x,s) e~sx/a
If we let
a = x/a
and let f(s)
With this notation Eq. (4.76) can be written as
Y(x,s) = f(s)e"as
UPon consulting the tables of transforms, we find
y(x,t) = L"1f(s)efltS = F(t - a)u(t -a.)
Since
F(t) - L"1 f (s) = AQSin &>t
and u(t) is the unit step function defined as
u(t) ='
1 t > a-
0 t < d
Therefore, the desired solution is
y(x,t)
(A si
°0
sin <u(t - «) t > a
t < a.
(4.76)
(4.77)
(4.78)
(4.79)
(4.80)
(4.81)
(4.82)
(4.83)
39
or
A sincj (t - x/a)o
t > x/au
y(x,t) = JO t < x/a (4.84)
This means physically that a point x cf the string s-fcays at
rest until the time t = x/a. Thereafter it undergoes motion identical
with that of the end x = 0 but lags behind it in time by the amount x/a.
The constant a is the speed with which the wave travels.
CHAPTER V
EVALUATION OF DEFINITE INTEGRALS
Certain types of definite integrals may be evaluated very easily
by the use of integral transforms. The Laplace transform is the most ex
tensively used integral transform, at the present time, to evaluate
definite integrals. We will present the general procedure by which defi
nite integrals are evaluated by means of examples. Many integrals may be
evaluated by the following theorem.
Theorem 5.1: If L f(t = F(s), then
rns) ds ■ /The proof of this theorem may be done in the following manner:'
Proof: By hypothesis we have
F(s) -f e"Stf(t) dt (5.2)J o
Therefore
r00 f(s) ds = r°D r00 e~stfu) dt dS (5.3)V o o J o
•00 00
f f e"Stf(t) ds dtJ o J o
provided that it is permissible to reverse the order of integration.
40
41
But we have
00 -ste ds = I (5.4)
/ds =rii£l dt (5.5)
o o
A critical examination of the validity of the above theorem pro
cedure reveals that it is possible to change the order of integration
involving infinite limits.*
To illustrate the use of the above theorem, let's consider some
examples.
Example 1: Evaluate the integral
dt where a > 0 (5.6)f sin at
Since
a
L sin at =
s + a
then by theorem 5.1
2 • -2 (5.7
/°° sin at dt r*0 a_
~ ~J '. s2 +sin_at_dt .f*" „ * ^ ds
a
Tan"1 sCD
= E (5.8)
2
o
*A discussion of this will be found in H.S. Carslaw, "Fourier Serives
and ingergal" Dover Publication, Inc., New York, 1951.
42
Example 2: Evaluate the integral
Jo t
(5.9)
Since
I e"at = 1/(b + a)
and
L e"bt b)
then
(5.10)
(5.11)
r e~at - e"bt dt =r fri_ _ _iJo t J° I s + a s +
Ids
ln(s + a) - ln(s + b)
s + a
s + b
GD
00
(5.12)
Another method of evaluating of integrals operationally depends
on the introduction of a parameter in the integrand. As an example, con
sider the following
Example 3; Evaluate
Q0
, -CD
x sin xt dx
a + x
(5.13)
Since
/ x sin xt dx _ / /J q o o ""/n •* ft2 2
a + x
e-st xsinjct dx dt
a + x
(5.14)
43
Changing the order of integration we have
J a + x ■
—Ste sin xt dt dx
o (a + x ) (s + x )
Resolving the above into partial fractions we have
2
s
^r 2 2 1i r f * *
27 2 2 2^2-a^o [s +X a+x J
dx (5.16)
i / -l -l Mi~2 ^; I sTan x - aTan x )
s - a * J
(Its _ E!L
2 2
Since
/ x sln xt dx = if x sin xt dxa2+x2 JO ~ "
Then
C x sin xt dx _ r_JL_"|J 2 2 s + a I" ^qd a + x L J
Taking the inverse transform, we have
Therefore
2^2a + x
-OD
2^- cd a + x
OD
(5.18)
-1 n = rte~at (-5.19)L TTT
00 xsin xt dx _ rfc-at (5#20)
44
Theorem 5.2: If the integral
f f(t) dt
o
,0D t
converges, then the integral/" e" f(t) converges uniformly with respect
to s in the close interval O.< s ^ s^ for any real value of s1> 0
That is
cd op
I f(t) dt = lim j e f(t)dt = lim F(s) = a constant.
We will not give a proof of the theorem; but the theorem can be
illustrated by some examples.
Example 4: Consider the integral
I = F J (t) dt (5.21)
Since the integral converges, then by theorem 5.2, we can write
' J (t) dt = linif e"StJ (t) dt (5.22)0 s-*0^o o
From the tables of transforms we have
lim f° e~StJ (t) dt = lim _ X =1 (5.23)
Example 5; Evaluate the integral
r sin xt dx (5.24)J o . /——
45
We know that
2,4*_ + X_
3! 5!
where
However
Hence from
2r,
4.
sin
X
Eq.
=
X
(5
2
■(.26),
1 -
we
3!
have
(5.26)
* <5-27>
therefore
sin x = Y 2 J, (x) (5.28)
Since I converges, then by theorem 2, the following integral
converges:
OD QD
sinjct =[»[e-SK Jx (xt)* -V2j ^dx (5.29)
From the tables of transforms, we have
rvTJ
Now Eq. (5.30) can be written as
L (s2 + t^ J(5.32)
4b
Applying theorem 5.2 to Eq. (5.32), we have
od I 2 2 k;in xt dx = J llm (s + t K - s.I = r sin xt dx = rjj-
(5.33)
Remarks concerning Theorem 5.2
Uniform convergence on the interval 0 ^ x<co also means that
F(s) (the Laplace transform of a function f(t) is continuous in the in
terval 0 < x < cd , that is, for all values of s therein). It must not
be inferred however, that the continuity of F(s) implies uniform conver
gence of the integral inO<s<. oo . For instance as s -* 0, F(s) may be
continuous while the integral diverges. Consider the following example
llm rCDe"Stcos t dt - lim a = 0
J
ButCD
/cos t dt
o
diverges.
The integral diverges because f(t) = cos t, sin t are alternating
functions with a constant amplitude as t-»od. The function does not tend
to a limit as t-*<D , so
r cos t dt
oscillates finitely, that is, it is bounded and indefinite.
CHAPTER VI
APPLICATION TO ORDINARY NONLINEAR DIFFERENTIAL
EQUATIONS
For nonlinear differential equations there are no general methods
which are applicable to all cases and no solutions exist in closed form; in
general the solution cannot be written in terms of the classical- trans-
cedentals. We must be satisfied with approximate methods. Although the
Laplace transform is not immediately applicable in the previous way, since
it is a linear transformation and it particularly suited for solving
linear problems, it can, nevertheless, be useful in approximate methods.
Many of the analytical methods of practical importance are based
on attempting to find a solution as a combination of well-known tabulated
mathematical functions. It is recognized that an exact solution probably
cannot be found, but an approximate solution of sufficient accuracy may be
possible. While details of various methods differ, most of them are rather
similar and follow basically the same format. One part of the differential
equation is a linear equation that is simple enough to allow an exact solu
tion. The other part contains any terms that are difficult to handle and
will usually involve the nonlinear terms of the equation, and perhaps other
terms as well. The linear equation is solved so as to give the zero-order
or generating solution. This generating solution is then employed in some
way with the nonlinear terms of the original equation to produce first-
order corrections terms. These corrections terms are then combined with
47
48
the generating term to yield a first-order corrected solution, which is
an approximate solution of the original equation. The exact form of the
correction terms depend upon the particular details of the method being
employed.
If the degree of nonlinearity is sufficiently small, a single
application of this process, yielding a first-order corrected solution, >
may give sufficient accuracy. If the degree of nonlinearity is greater,
it is sometimes possible to obtain better accuracy by applying the method
a second time, so as to yield a second-order correction. Further repeated
applications of the method are possible theoretically, but practically the
mathematics is usually too complicated as opposed to the small increase in
accuracy obtained. Actually, an important uncertainty inherent in methods
of this sort is the error in the solution which they yield. It is not
always a simple matter to determine the error.
Since the determination of the error of the approximate solutions
to nonlinear differential equations is not readily done, we will not con
sider it her. Our concern is to demonstrate that the Laplace transforma
tion can be employed to nonlinear differential equations, and that the
approximate solutions obtained are as sufficient as those obtained by the
traditional methods.
6.1 Solution by Nonlinear Integral Equations
A method for solving nonlinear differential equations based on the
theory of nonlinear integral equations will now be described. This is an
imperative operational method, devised by Pipes, is intimately related to
Laelsco nonlinear integral equation.*
We illustrate the procedure by using the second order nonlinear
*This is a Volterra's nonlinear integral equation.
49
equation in operator form
2(0)x(t) + f x(t), x'(t), ... = e(t) (6.1.1)
here D = d/dt. The term X(D) is the linear part of the differential
equation, while f rx(t)| is the nonlinear part. Let us suppose that
s(0) = x'(0) = 0 (6.1.2)
although the problem can be anlayzed in cases where the initial conditions
are non-zero.
In order to formulate Eq.(.6.1»I)as an integral equation, let the
Laplace transforms be introduced.
L x(t) = I e s(t) dt = X(s) (6.1.3)
J oL e(t) = E(s) (6.1.4)
£ f [x(t)] = G(s) (6.1.5)
With this notation and initial conditions (6.1.2), Eq. (6.1.1) can be
written as
Z(s)X(s) + G(s) - E(s) (6.1.6)
XT/ N E(s) G(s) ,, ,
x(s) = zTsT " TTsT (6*1
or
For simplicity, we introduce the notation
H(s) = 1/Z(s)
With this notation Eq. (6.1.6) may be written in the following form:
X(s) - E(s)H(s) + G(s)H(s) (6.1.8)
If we let
L"1 H(s) = L"1 1/Z(s) - h(t) (6.1.9)
then x(t) is obtained by taking the inverse transform of Eq. (6.1.8).
Thus
x(t) = L"1 E(s)H(s) - L"1 G(s)H(s) (6.1.10)
50
It v/e make use of the convolution theorem and apply if. to both members of
(6.1.10), the following results are obtained:
t t
x(t) = j h(t - u)e(u) du -J
o o
which is a nonlinear integral equation for s(t) of Volterra type. With
f(x) = 0, the nonlinear component is absent so that
t
xQ(t) = f h(t - u)e(u) du (6.1.12)
o
and in this notation Eq. (6.1.11) may be written in the form
t
x(t) = x (t) - f h(t - u)f [x(u)] du (6.1.13)
o
Laelsco hes shown that thelimit of the infinite sequence of functions
t
x (t) - f h(t - u)e(u) duo J
o
t
X,(t) = x (t) - f h(t - u)f [x (u)] dux o J ,
° (6.1.14)
t
x (t) - f h(t - u)f Fxn(u)l du
is the solution of the integral equation (6.1.13). That is
x(t)= limxn(t)
is the desired solution.
The investigation of the convergence of the sequence (6.1.14)
is given in the treatise of Laelsco. The results of this investigation
51
indicated that the sequence converges rapidly in application of this
method of solution of special cases arising from physical problems.
A convenient operation procedure that has considerable utility
in the calculation of the sequence x (t) for Eq. (6.14) is based on
finding the inverse Laplace transform of Eq. (6.1.10) rather than Eq.
(6.1.14). From Eq. (6.1.10) we have
T1x(t) - L"1 X(s) = L"1!^ - I (6.1<16)
whereupon the sequence x (t) is defined by
Xo(t)
L"1 [Lf [xn(O]l
\ X(s) )xn+l(t) + Xo(t) "L" JLf [Xn(t)jl (6.1.17)
Formally, of course, the sequence *n(t) obtained from Eq. (6.1.17) is
the same as that obtain from Eq. (6.1.14). Practically the sequence is
more easily computed form (6.1.17) because the table of Laplace trans
forms greatly assist in the analysis.
As an illustration of the procedure consider the following exam
ples:
Example 1.1; Consider the following nonlinear differential equation
(a + 3cx2) dt + bx = e (6.1.18)dt
with the condition x(0) = 0
Eq. (6.1.18) can be written in the form of (6.1.1) as follows
dx^h . d («?> - e (6.1.19)*+b + cT
52
In this case we let
Z(D) j- OD + b, f(x) = c-^- (x3), and D = d/dt (6.1,20)
As aresult
H(s) = 1/Z(s) = l/(as + b) (6.1.21)
hence
h(t) = L"1 H(s) = -e-kt (6.1.22)a
where k = b/a.
The integral equation specified by x(t) is
k t
x(t) - x (t) - ~ f e"k(t"u)Dx3(u) du (6.1.23)o a J
o
and the operational form is
(6.1.24)
where
x (t) - L~3 / 2—V = e (1 - e"kt) (6.1.25)\as + b) b
The next approximation for x(t) is
r 3x.(t) = £(1 - e"kt) + cL"1 L DX ovw| (6.1.26)1 b I as + b
Now with
3 3
then we must calculate
Dx (t) = e (ke- - ke + ke )o —r
53
Carrying out the indicated operations with the aid of the tables
of Laplace transforms, the following results are obtained:
x (t) = e(i - e~kt) - ce3(3kte"k1: - 2-kt + 6e"2kt _ 3 (6.1.28)1 b ab 2 2
The higher approximations may be obtained ny computing more func
tions of the sequence (6.1.24). If the coefficients of the nonlinear
terms are small, the sequence converges rapidly and the second or third
approximations usually give the accuracy required.
Example 1.2; Consider the following nonlinear differential equation
y" + (2 + ay2)y' + (1 + by2)y =0 (6.1.29)
subject to the initial conditions
y(0) = 1, and y'(0) - c (6.1.30)
We can write Eq. (6.1.29) in the following form:
(y" + 2y' + y) + (ay'y2 + by3) = 0 (6.1.31)
where in this case we let
Z(D) = D2 + 2D + 1 (6.1.32)
f(x) = ay'y2 + by3 , (6.1.33)
and D - d/dt
As a result
H(s) = c/Z(s) = c/(s2 + 2s + 1) = c/(s + I)2 (6.1.34)
and hence
h(t) « L"1 H(s) - cte-t (6.1.35)
The nonlinear integral equation that satisfied y(t) is
y(t) = yo(t) - c t (t - u)e"(t " U) [ay'(u)y2(u) + by3(u)] du
(6.i;36)
54
and the operational form is
y(t) = yQ(t) -
where in this case
L"1 [Uay'y2 + by3)]2
(s + 1)L J
-tyQ(t) - h(t) = cte
The r.ext approximation for y(t) is
cet"t - L"(ay'oy3o + by3)"
(s + I)2
Now with
y2(t) = c2tV2t, y3(t) = c3tV3t, and
y«(t)
we can calculate
t)e-t
ay •y2 + by3 = c3(at2 + (b - a)t3)e3N -3t
' o' o o
and then to form its Laplace transform:
(6.1.37)
(6.1.38)
(6.1.39)
(6.1.40)
(6.1.41)
(6.1.42
ay 'y2 + byo o
(s + 3) (s + b) J(6.1.43)
Hence we obtain the second approximation of the image function
1 S2ac 60? - a)
(a + I)2 (s + l)2(s + 3)3 (s + i)2(s + 3)4
(6.1.44)
where
Y(s) = L y(t) e"Sty(t) dt (6.1.45)
To find the corresponding original function y.(t), we make use
of the convolution theorem. We can write
2ac" 2ac"
(s + l)2(s + 3)3 (s + I)2 (s + 3)3
(6.1.46)
and
6(b - a)c = 6(b - a)c"
(s + l)2(s + 3)4 (s+1)2 (s + 3)4
Since (6.1.47)
L"1 te-t
(s + 1)'
,-1 1-3t
(s + 3)1
(s
l/6tV3t
We have by the convolution theorem
2ac
(s + l)2(s + 3)3
. u)e-(t -
ac f (u2t - u3)e"2u"tdu (6.1.48)Jo
6(b - a)c
(s + 1) (s +
_ u)e"<t -u) du
c3(bC , 3 4. -2u - t
-a) / Cu t - u )eJ a
du
(6.1.49)
56
Combining Eq. (6.1.18) and (6.1.49), we find the original function to be
y (t) = cte - c b f \\n t - u t) - u - u )e du
J o
(6.1.50)
Integrating Eq. (6.1.50) by parts and simplifying, we find the
first-order corrected solution to be
(6.1.52)
The continued application of this process obviously produces a sequence
of odd exponential functions multiplied by polynomials.
6.2 Solution by Power Series
Another powerful method of determining the solutions to certain
nonlinear differential equations will be given here. The method presented
is an operational adaptation of the one developed by Linsted and Liapound-
off. The solution of nonlinear differential equation is obtained as an in
perturbation theory by introducing a power series of the parameters and
then applying the Laplace transform to find each term. This operational
method, based on the Laplace transform, for solving certain problems has
been studied extensively by Pipes.1 The introduction of this operational
process brings to bear extensive tables of transforms thereby reducing
the algebraic labor involved in obtaining practical solutions.
This method can be illustrated by the following equation:
lL. A. Pipes, Operational Methods in Non-linear Mechanics (New
York: Dover Publications, 1965).
57
Example 2.1: Consider the following nonlinear differential equation
x"(t) +o>2x(t) +Xx2(t) = A(t) (6.2.1)
This equation is used in the theory of seismic wage by Nogaoka. The
effect of the point action of two simple harmonic forces will be con
sidered by letting
A(t) = a1cos<Jlt + a2cos£t)2t (6.2.2)
where a, a , a , o>» (O-,, &>2» and are constants subjected to the initial
conditions
x = a\I t = 0 (6.2.3)
x1 = 0)
If we let
y(s) = f00 e"Stx(t)dt (6.2.4)
and to solve (6.2.1) we let
n=o
where the x.'s, whose transform we denote by y^s), are to be determined.
If Eq. (6.2.5) is squared, we have
x xn \2xn \2 = x2 + 2x x. X + x2 \ + 2(x + x )x X2 + ..•nJ o ol . o ± £■
b /
(6.2.6)
Taking the Laplace transform of Eq. (6.2.1) we find
o 2 2*■ r_\ / n\ — l/r>\ J. .. / r,\ -L. T v
s y(s) - sx(0) - x'(0) + y(s) + L x =
~2 2 2 . .2s +tt>i s +6J2
58
Upon substituting the initial conditions (6.2.3), and simplifying, then
/ •» as . ais a2s L x2o 9 2 22 22
s + oj (s + <j )(s + 6>i) (s + cu )(s
The transform of Eq. (6.2.5) is
y(s) = ^ y (s)Xn . (6.2.8)
X nn=o
Upon substituting Eq. (6.2.5) and (6.2.6) into Eq. (6.2.7), we find that
n=o
2 2s + o>~
n _JL_+ ,
-a2+ft|2 (S2 + cu2)(s2+coj) (S2 + .o>2)(s2+ c2
"- (6'2'9)
If we onlv consider powers of X no higher than the third, then when is
raised to the ze.ro, first, second, and the third powers we have
3.1 S
v
2(6.2.12)
s2+o>2
(s) = L [2(xQ + ,1)»2 + 2] (6>2a3)
3 2 2s + io
With the y 's thus identified that the system has the solution
x = N X_A. - L / 'AZ- n ^- nn=o n=o
Now we shall use the notation
X(a) - s> -
s2+a2 (6.2.15)
then Eq. (6.2.10) may be written in the form
y (s) = aeT(w) + a1sT(co)T(co1) + a2sT(co)T(u)2) . (6.2.16)
Inverting we have
x (t) = acos cjt + a (cos^t = cos coit) a2(cos ut - cos cc2t) (6.2.17)o _1 +
2 2 2 _ 2o>i ~ to o)_ ~ w
If we let
u ~ u>i id ~ U2
A - (a + Ax + A2)
then ve can rewrite Eq. (6.2.17) as
x (t) « Acos ut + A.cos u.t + A cos o)2t. (6.2.18)o j. i t
This represents the first approximation of the solution of Eq. (6.2.1)
subjected to the conditions (6.2.3). The transformed second approximation
is given by Eq. (6.2.11), and may be written in the form
yl(s) -TUi^ (6-2-19)2
Now we must calculate xo
x2 = A2cos2 cot + 2AA cos cotcos u^t + 2AA cos a)tcos a>2t +o x
A2cos2 cot + 2A1A2cos u^cos o,2t + A2cos2 U2t (6.2.20)
60
Then we find the Laplace transform to be
L x2 = !sA2(sT(2o$ + 1) + ^(sTCo.) + 1) + %A2(sT(2u)0) + 1) +o 112^
sAA, T(o) + a),) +■ T(o) - u),) + sAA, T( u+ n,.) + 'f^ - u ) +
sA.A_ T(o). + U/.) + T(ojj - o>2)
Now upon substituting Eq. (6.2.21) into Eq. (6.2.19) and computing the
inverse transform by means of the tables, and simplifying the results,
we find that _
o 2 9 I A A, + A9 A.
A + Al + V + C°S Ut 1 -2 + — V- + 9 912 . I 3to2 2oj2 (2o,2 _ ^2)
A2 Ml Ml M22 2 2
( T) () ) (w, + Tum^) (wi
0) 2 - o0l -h,2>2 ^-(^ - o)^2' 6u)
A2
"r
2 cos
2 2Al cos ^l* + A2 cos 2oi2t +
- 2^2) (8w2 - 2o)2) o)2
AA ' x *- ' AAOcos(u+u ) + 1
1 - 2'2
cos (o)+o)2)t+ ^2 cos( o)-o)2)t: + L2.
+ "2^ ~ 0) J
cos U, + o^)t + \_2 cos
is -2
- 0)
A second correction x (t) is obtained in a like manner.2
CHAPTER VII
APPLICATION TO INTEGRAL EQUATIONS
OF THE CONVOLUTION TYPE
An integral equation is an equation in which the function to be
determined appears under an integral sign. The importance of integral
equations in mathematical applications to physical problems lies in the
fact that it is usually possible to reformulate a differential equation
together with its boundary conditions as a single integral equation.
Integral equations of most frequent occurence in practice are convention
ally divided into two classifications, First, an integral equation of
the forts
b
#x)y(x) « P(x) -A / K(x,t)y(t) dt (7.1)'*' a
where <f>, F, and K are given function and A, a, and b are constant s is
known as a Fredhold equation. The function y(x) is to be determined.
The given function K(x,t), which depends upon the current variable as well
as the auxiliary variable t, is known as the kernel of the integral equa
tion. If the upper limit of the integral is not a constant, but is
identified instead with the current variable, the equation takes the form
<£(x)y(x) = F(x) + A r K(x,t)y(t) dt
a
and is known as a Volterra equation.
61
-x
(7.2)
62
When <p 4 0, the above equation involves the unknown function
y(t), both inside and outside the integral. In the special case when
<p = 0, the unknown function appears only under the integral sign, and the
equation is known as a Fredholm integral equation of the first kind, while
in the case when <p = 1 the equation is said to be of the second kind.
Equations (7.1) and (7.2) have one thing in common; they are both
linear integral equations. That is, the function y enters the equation
in a linear manner so that
ra K(x,t) ["^(t) + c2y2(t) ]J vb
b
= ci/K(x,t)y1(t) dt + c2 TK(x,t)y2(t) dt (7.3)
* a a
If the integral were replaced by the more general
rbK(x,t,y)(t) dt <7-4>
one would call the quation nonlinear. Equation (6.1.11) or
TK(x,t)y2(t) dt <7'5)
a
are typical examples of such operators.
We will now discuss certian types of integral equations that can
be solved by the use of the Laplace transformation.
7.1 Integral Equations of the Convolution Type
A special integral equation of importance in application is Vol-
terra's integral equation of the second kind with a different kernel.
The equation is of the form
63
t
Y(t) = F(t) + T K(t - u)Y(u) du (7.1.1)
o
where K(t - u) and F(t) are known functions and the function Y(t) is to
be determined. This equation is known in practice as the Convolution
type, which can be written as
I(t) = F(t) + K(t)*Y(t) (7.1.2)
where * indicates the operation of convolution. The unknown function
Y(t), and hence the solution of (7.1.1) may be readily found by the use
of the Laplace transformation. To solve (7.1.1) we shall introduce the
following transforms:
L F(t) =■■ f(s) (7.1.3)
L Y(t) = y(s)
L K(t) - k(s)
Than making use of the convolution theorem, we have
L f K(t - u)Y(u) = k(s)y(s) (7.1.4)
o
Now upon taking the Laplace transformation of Eq. (7.1.1), assuming that
both f(s) and k(s) exist, we find
y(s) = f(s) + k(s)y(s) (7.1.5)
This equation may be solved for y(s) in the form
V(S) = f(s) <7-1'6)1 - k(s)
The required solution may be found by inversion, therefore
L"1 f(s) £7.1.7)
1 - k(s)
In the form (7.1.6) y(s) cannot be immediately transformed back into the
original function Y(t). We can write y(s) in the form
y(s) = f(s) + k(s) f(s) (7.1.8)
1 - k(s)
and it can be shown that an original function Q(t) always corresponds
to the function
q(s) = k(s) (8.1.9)
1 - k(s)
and thus (7.1.10) can be transformed in the original function, giving
Y(t) = F(t) + Q(t)*F(t) (7.1.10)
This solution, when it is written in the form
F(t) = Y(t) + (-Q)*F (7.1.11)
has the same form as the original integral equation except, that the
roles of Y(t) and F(t) are now interchanged and in place of the kernel
K(t) we have the 'reciprocal kernel' -Q(t).
We will now consider some examples to illustrate this procedure.
Example 1.1; Solve the integral equation
t
Y(t) = at +/ sin(t - u)Y(u) du (7.1.12)
The integral equation can be written in the form
Y(t) = at + Y(t)*sin t (7.1.13)
Taking the Laplace transformation, and using the convolution theorem we
find that
y(s) = a/s2 + y(s)/(s2 + 1) (7.1.14)
Solving, we obtain
y(s)=a[_ + \ (7.1.15)V 2 4 /\s s /
65
Upon consulting tha table of transforms, we find
Y(t) =. a(t + |t3) (7.1.16)6
which can be verified directly as the solution of the integral equation
(7.1.12),
Example 1.2: Solve the following nonlinear equation:
Y(t) « h sin 2t +f Y(t - u)Y(u) du (7.1.17)^o
We can write the integral equation in the form
Y(t) = h sin 2t + Y(t)*Y(t) (7.1.18)
Then taking the Laplace transform, using the convolution theorem we find
that
y(s) - _1 . 2 (7.1.19)82 + 4 +(y(s))2
Solving, we obtain
(y(s))2 - y(s) + 1 = 0 (7.1.20)s2 +"4
We obtain the image solution
Y(s)
"9 ~ 2 I
4 \H (7.1.21)2 2 x ^/^ + 4
-1 + 1
Thus
1 ( s2 + 4 - s \ (7.1.22)
2 VWT7
66
and
y(s) - i / Vs'+ 4 + s \ (7.1.23)
From Eq. (7.1.22) we find the solution
Y(t) = L^yis) = J^t) (7.1.24)
Then Eq. (7.1.23) can be written as
y(a) - 2 1 (7.1.25)
's2 + 4
^ Vs+ 4 /
(7.1.26)
Hence a second solution is
Y(t)■- 5(t) - Jx(2t) (7.1.27)
where 5 (t) is the Dirac Delta function which vanishes when t * 0, but
is indinitely greater at t < 0. Therefore the solution (7.1.24) is con-
tinuous and bounded for t <^ 0.
Example 1.3; Convert the following differential equation to an integral
equation, and then solve the integral equation:
Y« - Y« - Y - 0 C7-1'28)
subject to the initial conditions
Y(0) = 0 and Y'(0) = 1 (7.1,29)
Integrating both sides of Eq. (7.1.28) form 0 to t, we have
f [y"(u) - Y'(u) - Y(u)J du = 0 (7.1.30)
67
or
Y'(t) - Y'(0) - Y(t) + y(0) - J Y(u) du = 0 (7.1.31)
o
Considering the initial conditions we have
Y'(t) - Y'(0) - Y(t) - f Y(u) du (7.1.32)J o
Integrating a second time from 0 to t, we have
Y(t) - Y(0) - Y'(0)t - J Y(u) - / (t - u)Y(u) du (7.1.33)
o o
Considering the initial conditions again, we find
Y(t) - t - J Y(u)du - J (t - u)Y(u) du (7.1.34)
o o
Therefore we obtain the following integral equation
Y(t) = t +/ Y(u) du = / (t - u)Y(u) du (7.1.35)o o
We can write the above equation in the form
Y(t) = t + Y(t) + t*Y(t) (7.1.36)
Taking the Laplace transform we find
y(s) = 1/s2 + y(s)/s + y(s)/s2 (7.1.37)
Solving, we obtain
y(s) 1 (7#1s2 - s - 1
We can write Eq. (7.1.38) as
( ) 1 ^7*]' 2 5(s - h)2 - I
68
Hence
y(s) =V5 p. vg -IL (s - h)2 - ( 5/2)2J
(7.1..40)
If we let
a = -v/5/2 and b =
then Eq. (7.1.40) becomes
yOO-7 r s _] (7.i.ii)Z J
7 r s _]1 (s - br + aZ J
Then consulting the tables of transforms, we obtain the following solution
Y(t) = -ebtsin ata
therefore
| |-t (7.1.42)
7.2 Abel's Integral Equations
An important integral equation of the convolution type is Abel s
integral equation, which takes the following form:
t
F(t) = f (t - u)1^) du (7.2.1)
o
where n is constant such that 0 < n < 1.
Equation (7.2.1) is classified as a Volterra's integral equation of
the first kind with a different kernel which is a special case of Eq.
(7.1), and can be written in the form
,t
F(t) - f K(t - u)Y(u) du (7.2.2)
o
Where the kernel K(t - u) is known and F(t) is a known function, and it is
69
required to determine the unknown function Y(t).
The transformed equation of F,q. (7.2.2) is
f(s) = k(s)y(s) (7.2.3)
Its solution is
y(s) - f(s)/k(s) (7.2.4)
In order to obtain a solution to Abel's equation from (7.2.3) we
have
L K(t) = L"1^11 = s11"1 V(l - n) = k(s) (7.2.5)
we obtain the image equation for y(s)
- n) \ y(s) - f(s)-n
with the solution
y(s) « f _f(s)
Since
n
n-1
the corresponding original function is
Y(t)
From the well-known formula
n-l*F(t)
V(n)
sin nft
n
(7.2.6)
(7.2.7)
(7.2.8)
(7.2.9)
(7.2.10)
Then, by application of the convolution theorem, we obtain
sin nn d Iit dt J
F(u) du (7.2.11)
(t - u)n-1
70
which is the solution of Abel's integral equation where 0 < n -<1.
We will not consider a special case of Absl's equation:
Example 2.1; Solve the integral equation
t V 2Y(u)(t - u) 2du = 1 + t + t
The equation can be written in the following form:
i 2
Y(t)*tT* - 1 + t + t
Than taking the Laplace transform, we find
Y(s) Vfls) =L +h + 2-
(7.2.12)
(7.2.13)
(7.2.14)
Sovoing we have
y(a) - 4- * 4- * -i-
Inverting, we find that
Simplifying
Y(t) = 8t
2t3/2
(7.2.15)
(7.2.16)
(7.2.17)
APPENDIX A
TABLES OF LAPLACE TRANSFORMS
This Appendix contains common Laplace transforms which were implicitly
encountered in this research. A table of transforms sufficiently exten
sive for most practical purposes is given by Pipes and Harvill.*
*L. A- ?ipes and L. R. Harvill, Applied Mathematics for Engineers andPhysicists. Appendix A, pp. 764-782.
71
t>, Table 1 Theorems
F(s).
F(s) = f f(t)e"St dt
o
f(t)
+ ioo
f(t)J_ f2ni J
F(s)eSt ds
-iCD
= Real
k
s
kF(s)
sF(s) -
s2F(s)
f(0)
- sf(O)
k
kf(t)
dt
A'or
F1(s)F2(s)
e"aSF(s)
f2(^)f1(t -v(dv
o ,
f(t - a)u(t - a) a - Real 0u(t) = Heaviside
unit step
lira F(s)
s rf(t)dt
dnF(s)tnf(t) n = integer
riE(x.s)df(xtt)
dx
OD
F(s) ds
72
CD
f(t) dt
t
APPENDIX B
TABLES OF TRANSFORMS
Table 2 Table of Transforms
s(
2s
2s
2s
1
1
+
s
1
-
s
a)
2a
2a
2a
i Ota .
" u(t) = h t - a u(t) = Heaviside step function
1 t a
/cd t = ° 5(t) = Dirac delta function
00 5(t)dt = 1
- oo
(0 t < a1 -as ' u(t - a) = <h t = a a = Reals (1 t > a
i. = t_n+1 ni
s n*
n n = 1, 2, 3, ..
n n = all values except
negative integers
n+1 »^n ' V(x) = gammas
1 e
s + a
-at
- a©
s + a
1/-. -at\—(1 - e )
ata
cos at
h ata
cosh at
2 2s - a
1) V(x) = gamma function
-at a _ real or complex
(s + a)'
-atte
-bt ., , , .2 2 e sin at(s + b) + a
s f b -bt— e cos at
2 2(s + b) + *
T(6>) sin cot T(o>)
|^T(cu) - T(3cu) sin3ojt
~-l - s2T(2(u) sin2 ojis
sT(a») cos bit
1 2 2— 1 + s I(2dj) cos cut2s
3
3sT(a>) t- sT(3<u) cos
J5 (A + B)T(A + B)
+(A - B)T(A - B) sin Atcos Bt
■f T(A + B) + T(A - B costAtcosBt
— T(A - B) - T(A + B) sinAtsinBt
Sf(a)T(b)
-a vs
2 v/rrt
cos
b
at
2
a
-
a
cos
2
-ae
bt
2/4t
2 2s +<u
2 j. 2s + a
- s)
2^2s + a
Ta
75
-ate
~\7«ft"
J (at)o
I x
1 - ert
2 s/t
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Churchi.ll, R. V. Fouries Series and Boundary Value Problems, New York:
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Doetsch. A Guide to the Application of the Laplace and Z-Transformations.
Springer, 1968.
Drake, Reuben C. Application of the Laplace Transformation to the
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Ford, Lester R. Differential Equations. New York: McGraw-Hill Book
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Harvill, L. R. and Pipes, L. A. Applied Mathematics for Engineers and
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Hoscstadt, Harry. Integral Equations. New York: John Wiley and Sons,
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